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This semester, I'm taking Calculus two. I'm working on a project for myself where I will need to solve the for a variable, inside a summation, such as "i" in the equation the one below:

$$ 0=-20+\sum _{n=1}^{\infty \:}\:\frac{2}{\left(1+i\right)^n} $$

Without telling me how to solve, is this A) possible B) something that would be explained in a Calc 2 course C) a hint, please? :)

Felix Marin
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findingmyway
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1 Answers1

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Without telling me how to solve, is this A) possible

I imagine you mean to solve for $i$ in the summation? It would be possible, in theory: just find a closed form for the summation. A similar example would be the below (assume $|x|<1$):

$$0 = -20 + \sum_{n=1}^\infty x^n$$

Say you wanted to solve for $x$ here; well, a priori, it seems very difficult, because there's infinitely many $x$'s. However, recall the form for a geometric series:

$$\sum_{n=1}^\infty x^n = \frac{1}{1-x}$$

Then the equation is just

$$0 = -20 + \frac{1}{1-x}$$

for which solving for $x$ is almost trivial.

Without telling me how to solve, is this B) something that would be explained in a Calc 2 course

I don't know if it would necessarily be explained in a Calculus 2 course, no; I don't recall it coming up in mine at least. However, figuring out identities, properties, and closed forms for certain series is a pretty central topic to Calculus 1 and 2; what you want to do is just an application of said work, really.

Without telling me how to solve, C) a hint, please? :)

Assuming your series converges, you can factor out a $2$, and then notice from exponent rules that this holds:

$$\sum_{n=1}^\infty \frac{1}{(1+i)^n} = \sum_{n=1}^\infty \left( \frac{1}{1+i} \right)^n$$

Then consider utilizing the geometric series.

PrincessEev
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    Wow! I wasn't expecting such a detailed response! I really appreciate your assistance! Thank you! – findingmyway Aug 08 '20 at 03:12
  • Ok question. Here is my new equation $20=\sum _{n=1}^{\infty }:\left(\frac{1}{1+x}\right)^n$

    If we use the identity $\sum _{n=1}^{\infty }:x^n=\frac{1}{1-x}$

    Then $x=\frac{1}{1+x}$ and $20=\frac{1}{1-\frac{1}{1+x}}$, with x equal to 1/19.

    If we put x=1/19 back into the original equation of $20=\sum _{n=1}^{\infty }:\left(\frac{1}{1+x}\right)^n$, the summation equals 19, not 20.

    Where has my math gone wrong here?

     – findingmyway Aug 10 '20 at 15:56
  • You must have made an arithmetic error in putting $x=1/19$ back into the original series; my by calculations it is the correct value. – PrincessEev Aug 10 '20 at 19:41
  • So I'm cheating by using Symbolab and then following the work. It seems the discrepancy comes from the starting index. When one uses a starting index of 0, the summation equals 20. Using the starting index of 1 (which I am) gets us to 19. – findingmyway Aug 10 '20 at 21:27
  • Ah, right, I didn't notice. But yeah, that starting index is important: for $|x|<1$,

    $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

    but, if you start at $n=1$, you have to subtract one from the right-hand-side:

    $$\sum_{n=1}^\infty x^n = \frac{1}{1-x} - 1 = \frac{1-(1-x)}{1-x} = \frac{x}{1-x}$$

     – PrincessEev Aug 10 '20 at 21:31
  • Got it. Thank you! – findingmyway Aug 10 '20 at 21:46